Find roots of polynomial $$ p(x) = x^4-x^2+\frac{1}{8} $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0.3827\\[1 em]x_2 &= -0.3827\\[1 em]x_3 &= -0.9239\\[1 em]x_4 &= 0.9239 \end{aligned} $$Explanation
Step 1:
Get rid of fractions by multipling equation by $ \color{blue}{ 8 } $.
$$ \begin{aligned} x^4-x^2+\frac{1}{8} & = 0 ~~~ / \cdot \color{blue}{ 8 } \\[1 em] 8x^4-8x^2+1 & = 0 \end{aligned} $$Step 2:
Polynomial $ 8x^4-8x^2+1 $ has no rational roots that can be found using Rational Root Test, so the roots were found using quartic formulas.
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